A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?
A) 2.0 N
B) 5.0 N
C) 1.0 N
D) 3.6 N
E) 3.0 N

Question

Quizplus · Accepted Answer

The force on a current-carrying wire in a magnetic field is given by:

$\vec{F} = I\vec{L} 	imes \vec{B}$

where $\vec{L}$ is the vector representing the length and direction of the wire segment, $\vec{B}$ is the magnetic field, and $	imes$ represents the cross product.

We can break up the wire segment into two parts:

1. From $x=-2.0$ m to $x=0$, the wire is parallel to the magnetic field and there is no force.

2. From $x=0$ to $y=3.0$ m, the wire is perpendicular to the magnetic field and the force can be found using the above equation.

The magnitude of the force is given by:

$F = ILB\sin	heta$

where $	heta$ is the angle between $\vec{L}$ and $\vec{B}$, which is 90 degrees in this case.

Using the given values, we get:

$F = (25	ext{ A})(3.0	ext{ m})(0.040	ext{ T})\sin(90^\circ) \approx 3.6	ext{ N}$

Therefore, the answer is D) 3.6 N.

A Segment of Wire Carries a Current of 25 a Along